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factorise f(x)=2x^3-x^2-13x-6

Sagot :

esterzan

Answer:

x1 = -½

x2 = -2

x3 = 3

Step-by-step explanation:

replace f (x) with 0: 0=2x³-x²-13x-6

moves the expression to the first member:

-2x³+x²+13x+6=0

rewrite the x² and 13x as a sum:

-2x³-x²+2x+x+12x+6=0

now pick up -x² from the expression:

-x²(2x+1)+2x²+x+12x+6=0

now pick up the x:

-x²(2x+1)+x(2x+1)+12x+6=0

now pick up 6:

-x²(2x+1)+x(2x+1)+6(2x+1)=0

finally pick up 2x + 1 and rewrite the x as a difference:

-(2x+1)(x²+2x-3x-6)=0 (+2x-3x is the x rewritten as the difference)

pick up x and -3:

-(2x+1)•[x(x+2)-3(x+2)]=0

pick up x + 2 and change the sign:

(2x+1)(x+2)(x-3)=0

now having this, we can solve the equations in x:

2x+1=0 ---> x=-½

x+2=0 ---> x=-2

x-3=0 ---> x=3

here are the three solutions:

x1 = -½

x2 = -2

x3 = 3

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